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If $x \geq \min\{y,w\}$ then $x \leq y+w$.

This is kind intuitive,very trivial. Or not? It's like triangular inequality, isn't it?

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2  
$x \leq y+x$ is just $0 \leq y$, no? –  J. M. Aug 14 '12 at 13:31
    
Now I edited...it´s $x\leq y+w$.My bad. –  HipsterMathematician Aug 14 '12 at 13:35
    
@MeAndMath It is still false as stated. See sebigu's example below. –  Derek Allums Aug 14 '12 at 13:37
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Every $x$ sufficiently big will disprove your conjecture. –  enzotib Aug 14 '12 at 13:46

1 Answer 1

up vote 3 down vote accepted

This is wrong. Take $y,w=1$ and $x=3$.

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Yes,yes,yes.If you considerate $y,w$ to be equal it´s wrong.Right.I forgot that possibility...what if $y\neq w$ ? –  HipsterMathematician Aug 14 '12 at 13:38
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Then take $y=0, w=1$ and $x=2$. –  Derek Allums Aug 14 '12 at 13:39
    
Take $y=2,w=3$ and $x=6$. –  sebigu Aug 14 '12 at 13:40
    
GREAT!THANKS IN ADVANCE! –  HipsterMathematician Aug 14 '12 at 13:40
    
No problem. Please mark this questing as solved. –  sebigu Aug 14 '12 at 13:42

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